Uniqueness of Neumann series

Question

Let $f$ be an entire function. Then there exist numbers $a_0,a_1,\ldots$, independent of $z$, such that $$f(z)=\sum_{n=0}^\infty a_n J_n(z),\quad \forall z\in\mathbb{C}$$ where $J_n$ is the Bessel function of the first kind of order $n$. This is called a Neumann series of $f$.

Question

Given $f$, are the numbers $a_n$ unique?

Thoughts

I took a look at Abramowitz and Stegun and Whittaker and Watson; in each book they mention the series but say nothing about the uniqueness of the coefficients. It would be nice if Neumann series turn out to be unique just as Taylor series of entire functions. I figured out that it is sufficient to prove that, for all $z\in\mathbb{C}$,

$$\sum_{n=0}^\infty a_n J_n(z)=0\implies \forall n\in\mathbb{N}:a_n=0;$$

Whittaker and Watson arrived at one possible integral representation of $a_n$ in terms of "Neumann polynomials". They essentially proved that if $a_n=\text{integral expression}$ then $f(z)=\sum_{n=0}^\infty a_n J_n(z)$, not the other way around.

This question has also been asked on MSE as can be seen at this link. According to the user Conrad in the comments, the uniqueness is known if the Neumann series are required to be uniformly convergent, but not known if they're only pointwise convergent.

Edit:

I'm thinking about using "orthogonality", i.e. $$\int_0^\infty \frac{J_{v+2l+1}(t)J_{v+2m+1}(t)}{t}\, dt=\frac{\delta_{l,m}}{2(2l+v+1)}$$ where $v+l+m\gt -1$ and then concluding that if $$\sum_{n=0}^\infty a_n J_n (z)=\sum_{n=0}^\infty b_n J_n(z),$$ then $$a_n=b_n$$ by the above orthogonality identity, but this requires me to justify switching $\sum_{n=0}^\infty$ with $\int_0^\infty.$

Please note that nothing about uniform convergence is assumed in this question; suppose that Neumann series only converge pointwise, nothing more.

Answer 1

Let me show that the convergence is uniform on balls, so that one can differentiate term by term and find the $a_n$ by evaluating at $z=0$, using that $J_n=c_nz^n+\dots$. It is more convenient to use modified Bessel functions $$I_n(z)=(-i)^nJ_n(iz)=\sum_{m=0}^\infty\frac{1}{\Gamma(m+n+1)m!}\left (\frac{z}{2} \right)^{n+2m}. $$ Then $f(iz)=\sum_{n} a_n J_n (iz)=\sum_n a_n i^n I_n(z)$.

Note that $I_n (1) \geq \frac{1}{n!2^n}$ and that $|I_n(z)| \leq \frac{1}{2^n n!}|z|^n e^{|z|^2/4}$.

Since the series converges at $x=1$ we get $|a_n| \leq Cn! 2^n$ and this gives uniform convergence on any ball or radius less than 1.

Answer 2

edit: This was meant to recall a first elementary but relevant fact that was not mentioned at all, that is orthogonality.

Bessel functions $\{J_n\}_{n\in\mathbb N_+}$ are orthogonal on $\mathbb R$ w.r.to the measure $dx/|x|$. So existence and uniqueness of the expansion, in the corresponding Hilbert space, is just an instance of the well-known general fact [edit: that is the Fourier expansion of an element of a Hilbert space in a given orthonormal basis, here the space spanned by the $J_n$'s].

[edit] in fact, since the degree of the lowest degree non-zero term of $J_n$ is $n$, the series makes sense even at the level of formal power series (for every sequence of coefficients), and you have uniqueness there (i.e. in the broader sense), and existence ($a_n$ are determined inductively from an infinite triangular system).

Answer 3

We can say a bit more: if a Neumann series converges point-wise, it also converges absolutely and in $\mathbb C[[z]]$, and the limit is an entire function.

Let $(a_n)_{n\ge0}$ be a sequence of complex numbers such that the Neumann series $$\sum_{n=0}^\infty a_n J_n(z)$$ converges point-wise on $\mathbb C$. Let $z\in\mathbb C$ and $r:=|z|$. We have (by discrete Tonelli) $$\sum_{k\ge0,\, n\ge0} \Big|a_n \frac{(-1)^k}{k!(n+k)!}\big(\frac z2\big)^{n+2k}\Big|=\sum_{n\ge0}|a_n| \sum_{k\ge0}\frac1{k!(n+k)!}\big(\frac r2\big)^{n+2k}.$$ The inner sum in $k$ writes $$\sum_{k\ge0}\frac1{k!(n+k)!}\big(\frac r2\big)^{n+2k}= i^{-n}\sum_{k\ge0}\frac{(-1)^k}{k!(n+k)!}\big(\frac {ir}2\big)^{n+2k}=|J_n(ir)|\le\frac{|J_n(2ir)|}{2^n}. $$ By assumption, the series $\sum_{n=0}^\infty a_n J_n(2ir)$ is also convergent, so $\big|a_n J_n(2ir)\big|\le M$ for some constant $M$. Then the above the double sum is majorized by $$\sum_{n=0}^\infty\big|a_n J_n(2ir)\big|\frac1{2^n}\le \sum_{n=0}^\infty\frac M{2^n}=2M<+\infty.$$

So the family of terms $\big\{a_n \frac{(-1)^k}{k!(n+k)!}\big(\frac z2\big)^{n+2k}: (n,k)\in\mathbb N^2\big\} $ is absolutely summable. Thus (by discrete Fubini) we can change order of summation and write $\sum_{n=0}^\infty a_n J_n(z) $ as a power series of an entire function $f(z)$; moreover, as already observed, $\sum_{n=0}^\infty a_n J_n(z) $ also converges in the sense of formal power series, and $a_n$ are inductively determined by the coefficients of $f$ .